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\begin{document}

\title{Complex Analysis}
\subtitle{Chapter 3. Analytic Functions as Mappings \\ Section 4. Elementary Conformal Mappings }
%\institute{SLUC}
\author{LVA}
%\date
%\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
%\date{ {2023年9月21日} }

\maketitle

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\begin{frame}{Contents 1-2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] Elementary Point Set Topology
\begin{enumerate}
\item[1.1.] Sets and Elements
\item[1.2.] Metric Spaces
\item[1.3.] Connectedness
\item[1.4.] Compactness
\item[1.5.] Continuous Functions
\item[1.6.] Topological Spaces
\end{enumerate}

\item[2.] Conformality
\begin{enumerate}
\item[2.1.] Arcs and Closed Curves
\item[2.2.] Analytic Functions in Regions
\item[2.3.] Conformal Mapping
\item[2.4.] Length and Area
\end{enumerate}

\end{enumerate}

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\begin{frame}{Contents 3-4}

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\begin{enumerate}
\item[3.] Linear Transformations
\begin{enumerate}
\item[3.1.] The Linear Group
\item[3.2.] The Cross Ratio
\item[3.3.] Symmetry
\item[3.4.] Oriented Circles
\item[3.5.] Families of Circles
\end{enumerate}

\item[4.] {\color{red}Elementary Conformal Mappings}
\begin{enumerate}
\item[4.1.] {\color{red}The Use of Level Curves}
\item[4.2.] {\color{red}A Survey of Elementary Mappings}
\item[4.3.] {\color{red}Elementary Riemann Surfaces}
\end{enumerate}
 
\end{enumerate}

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\begin{frame}{4. Elementary Conformal Mappings }

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\begin{enumerate}\itemsep1em

\item[1.] 
{\color{red}The {\color{blue}conformal mapping} associated with an {\color{blue}analytic function} affords an excellent visualization of the properties of the latter; it can well be compared with the visualization of a real function by its graph. 
}

\item[2.] 
It is therefore natural that all questions connected with conformal mapping have
received a great deal of attention; progress in this direction has increased our knowledge of analytic functions considerably. 

\item[3.] 
In addition, conformal mapping enters naturally in many branches of mathematical physics and in this way accounts for the immediate usefulness of complex function theory.

 
\end{enumerate}

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\begin{frame}{4. Elementary Conformal Mappings }

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\begin{enumerate}\itemsep1em

\item[4.] 
{\color{red}One of the most important problems is to determine the conformal mappings of one region onto another. }

\item[5.] 
In this section we shall consider those mappings which can be defined by elementary functions.
 
\end{enumerate}

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\begin{enumerate}\itemsep1em

\item[6.] 
When a conformal mapping is defined by an explicit analytic function $w = f(z)$, we naturally wish to gain information about the specific geometric properties of the mapping.

\item[7.] 
{\color{red}One of the most fruitful ways is to study the correspondence of curves induced by the point transformation. }

\item[8.] 
The special properties of the function $f(z)$ may express themselves in the fact that certain simple curves are transformed into curves of a family of well-known character. 

\item[9.] 
Any such information will strengthen our visual conception of the mapping. 

 
\end{enumerate}

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\begin{enumerate}\itemsep1em

\item[10.] 
Such was the case for mappings by linear transformations. 

\item[11.] 
We proved in Section 3 that a linear transformation carries circles into circles, provided that straight lines are included as a special case. 

\item[12.] 
By consideration of the {\color{blue}Steiner circles} it was possible to obtain a complete picture of the correspondence.
 
\end{enumerate}

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\begin{enumerate}\itemsep1em

\item[13.] 
In more general cases it is advisable to begin with a study of the image curves of the lines $x = x_0$ and $y = y_0$. 

\item[14.] 
If we write $f(z) = u(x,y) + iv(x,y)$, the image of $x = x_0$ is given by the parametric equations $u = u(x_0,y)$, $v = v(x_0,y)$; $y$ acts as a parameter and can be eliminated or retained according to convenience. 

\item[15.] 
The image of $y = y_0$ is determined in the same way. 
 
\end{enumerate}

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\begin{enumerate}\itemsep1em

\item[16.] 
Together, the curves form an {\color{blue}orthogonal net} in the $w$-plane. 

\item[17.] 
Similarly, we may consider the curves $u(x,y) = u_0$ and $v(x,y) = v_0$ in the $z$-plane. 

\item[18.] 
They are also orthogonal and are called the {\color{blue}level curves} of $u$ and $v$.

 
\end{enumerate}

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\begin{enumerate}\itemsep1em

\item[19.] 
In other cases it may be more convenient to use {\color{blue}polar coordinates} and study the images of {\color{blue}concentric circles} and {\color{blue}straight lines} through the origin. 

\item[20.] 
{\color{red}Among the simplest mappings are those by a power $w = z^\alpha$. }

\item[21.] 
We consider only the case of real $\alpha$, and then we may as well suppose that $\alpha$ is positive. 

\item[22.] 
Since
\begin{equation*}
\begin{aligned}
|w| &= |z|^\alpha \\
\mathrm{arg}\, w &= \alpha \, \mathrm{arg}\, z
\end{aligned}
\end{equation*}
concentric circles about the origin are transformed into circles of the same family, and half lines from the origin correspond to other half lines. 

 
\end{enumerate}

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\begin{enumerate}\itemsep1em

\item[23.] 
{\color{red}The mapping is conformal at all points $z \neq 0$, but an angle $\theta$ at the origin is transformed into an angle $\alpha\theta$. }

\item[24.] 
For $\alpha \neq 1$ the transformation of the whole plane is not {\color{blue}one to one}, and if $\alpha$ is fractional $z^\alpha$ is not even {\color{blue}single-valued}. 

\item[25.] 
In general we can therefore only consider the mapping of an angular sector onto another.

 
\end{enumerate}

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\begin{enumerate}\itemsep1em


\item[26.] 
The sector $S(\varphi_1,\varphi_2)$, where $0 < \varphi_2 - \varphi_1 \le 2\pi$, shall be formed by all points $z \neq 0$ such that one value of $\mathrm{arg}\, z$ satisfies the inequality
\begin{equation}
\varphi_l < \mathrm{arg}\, z < \varphi_2.
\label{eq-15}
\end{equation}

\item[27.] 
It is easy to show that $S(\varphi_1,\varphi_2)$ is a region. 

\item[28.] 
{\color{red}
In this region a unique value of $w = z^\alpha$ is defined by the condition 
\begin{equation*}
\mathrm{arg}\, w =\alpha\, \mathrm{arg}\, z
\end{equation*}
where $\mathrm{arg}\, z$ stands for the value of the argument singled out by the condition (\ref{eq-15}).
}
 
\end{enumerate}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}\itemsep1em

\item[29.] 
This function is analytic with the nonvanishing derivative 
\begin{equation*}
D e^{\alpha\log z} = \alpha \frac{w}{z}.
\end{equation*}

\item[30.] 
The mapping is one to one only if $\alpha(\varphi_2 - \varphi_1) \le 2\pi$, and in this case $S(\varphi_1,\varphi_2)$ is mapped onto the sector $S(\alpha\varphi_1,\alpha\varphi_2)$ in the $w$-plane. 

\item[31.] 
It should be observed that $S(\varphi_1 + n\cdot 2\pi, \varphi_2 + n\cdot 2\pi)$ is geometrically identical with $S(\varphi_1,\varphi_2)$ but may determine a different branch of $z^\alpha$.

 
\end{enumerate}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}\itemsep1em

\item[32.] 
{\color{red}Let us consider the mapping $w = z^2$ in detail. }

\item[33.] 
Since $u = x^2 - y^2$ and $v = 2xy$, we recognize that the level curves $u = u_0$ and $v = v_0$ are {\color{blue}equilateral hyperbolas} with the diagonals and the coordinate axes for asymptotes. 

\item[34.] 
They are of course orthogonal to each other. 

\item[35.] 
On the other hand, the image of $x = x_0$ is $v^2 = 4x_0^2(x_0^2-u)$ and the image of $y = y_0$ is $v^2 = 4y_0^2(y_0^2 + u)$. 

 
\end{enumerate}

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\begin{enumerate}\itemsep1em

\item[36.] 
Both families represent parabolas with the focus at the origin whose axes are pointed in the negative and positive direction of the $u$-axis. 

\item[37.] 
Their orthogonality is well-known from analytic geometry.

\item[38.] 
The families of level curves are shown in Figs. 3-5 and 3-6.

 
\end{enumerate}

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\begin{figure}[ht!]
\centering
\includegraphics[height=0.75\textheight, width=0.45\textwidth]{figure-3-5.png}
%\caption{Fig. 3-9. Mapping by $z=\frac{1}{2}(w+w^{-1})$ }
\end{figure}

\end{frame}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{figure}[ht!]
\centering
\includegraphics[height=0.75\textheight, width=0.45\textwidth]{figure-3-6.png}
%\caption{Fig. 3-9. Mapping by $z=\frac{1}{2}(w+w^{-1})$ }
\end{figure}

\end{frame}


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\begin{frame}{4.1. The Use of Level Curves. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}\itemsep1em

\item[39.] 
For a different family of image curves consider the circles $|w - 1| = k$ in the $w$-plane. 

\item[40.] 
The equation of the inverse image can be written in the form
\begin{equation*}
(x^2+y^2)^2 = 2(x^2-y^2) + k^2 -1
\end{equation*}
and represents a family of {\color{blue}lemniscates} with the focal points $\pm 1$.

\item[41.] 
The orthogonal family is represented by
\begin{equation*}
x^2 - y^2 = 2hxy + 1
\end{equation*}
and consists of all {\color{blue}equilateral hyperbolas} with center at the origin which pass through the points $\pm 1$.

 
\end{enumerate}

\end{frame}

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\begin{enumerate}\itemsep1em

\item[42.] 
{\color{red}In the case of the third power $w = z^3$ the level curves in both planes are {\color{blue}cubic curves}. }

\item[43.] 
There is no point in deriving their equations, for their general shape is clear without calculation.

\item[44.] 
For instance, the curves $u = u_0 > 0$ must have the form indicated in Fig. 3-7. 

\item[45.] 
Similarly, if we follow the change of $\mathrm{arg}\, w$ when $z$ traces the line $x = x_0 > 0$, we find that the image curve must have a loop (Fig. 3-8). 

\item[46.] 
It is a {\color{blue}folium of Descartes}.

\end{enumerate}

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\begin{figure}[ht!]
\centering
\includegraphics[height=0.75\textheight, width=0.75\textwidth]{figure-3-78.png}
%\caption{Fig. 3-9. Mapping by $z=\frac{1}{2}(w+w^{-1})$ }
\end{figure}

\end{frame}


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\begin{enumerate}\itemsep1em


\item[47.] 
{\color{red}The mapping by $w = e^z$ is very simple. }

\item[48.] 
The lines $x = x_0$ and $y = y_0$ are mapped onto circles about the origin and rays  of constant argument. 

\item[49.] 
Any other straight line in the $z$-plane is mapped on a {\color{blue}logarithmic spiral}. 

\item[50.] 
The mapping is one to one in any region which does not contain two points whose difference is a multiple of $2\pi i$. 
 
\end{enumerate}

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\begin{enumerate}\itemsep1em


\item[51.] 
In particular, a horizontal strip $y_1 < y < y_2, y_2 -y_1 \le 2\pi$ is mapped onto an angular sector, and if $y_2 - y_1 = \pi$ the image is a half plane. 

\item[52.] 
We are thus able to map a parallel strip onto a half plane, and hence onto any circular region. 

\item[53.] 
The left half of the strip, cut off by the imaginary axis, corresponds to a half circle. 

\item[54.] 
It is useful to write down some explicit formulas for the mapping. 

 
\end{enumerate}

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\begin{enumerate}\itemsep1em

\item[55.] 
The function $\zeta = \xi + i\eta = e^z$ maps the strip $-\pi/2 < y < \pi/2$ onto the half plane $\xi > 0$. 

\item[56.] 
On the other hand,
\begin{equation*}
w = \frac{\zeta -1}{\zeta +1}
\end{equation*}
maps $\xi > 0$ onto $|w| < 1$. 

\item[57.] 
{\color{red}
Hence
\begin{equation*}
w = \frac{e^z -1}{e^z +1} = \mathrm{tanh} \frac{z}{2}
\end{equation*}
maps the strip $|\mathrm{Im}\, z| < \pi/2$ on the unit disk $|w| < 1$.
}

\end{enumerate}

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\begin{enumerate}\itemsep1em


\item[1.] 
{\color{red}When faced with the problem of mapping a region $\Omega_1$ conformally onto another region $\Omega_2$, it is usually advisable to proceed in two steps.
}

\item[2.] 
First, we map $\Omega_1$ onto a circular region, and then we map the circular region onto $\Omega_2$.

\item[3.] 
In other words, the general problem of conformal mapping can be reduced to the problem of mapping a region onto a disk or a half plane. 

\item[4.] 
We shall prove, in Chap. 6, that this mapping problem has a solution for every region {\color{blue}whose boundary consists of a simple closed curve}.

\end{enumerate}

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\begin{enumerate}\itemsep1em


\item[5.] 
{\color{red}The main tools at our disposal are {\color{blue}linear transformations} and transformations by a {\color{blue}power}, by the {\color{blue}exponential function}, and by the {\color{blue}logarithm}.
}

\item[6.] 
All these transformations have the characteristic property that they map a family of straight lines or circles on a similar family.

\item[7.] 
For this reason, their use is essentially limited to regions whose boundary is made up of circular arcs and line segments. 

\item[8.] 
The power serves the particular purpose of straightening angles, and with the aid of the exponential function we can even transform zero angles into straight angles.


\end{enumerate}

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\begin{enumerate}\itemsep1em

\item[9.] 
{\color{red}By these means we can for instance find a standard mapping of any region whose boundary consists of two circular arcs with common end points. 
}

\item[10.] 
Such a region is either a circular wedge, whose angle may be greater than $\pi$, or its complement. 

\end{enumerate}

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\begin{enumerate}

\item[11.] 
If the end points of the arcs are $a$ and $b$, we begin with the preliminary mapping $z_1 = (z - a)/(z - b)$ which transforms the given region into an angular sector. 

\item[12.] 
By an appropriate power $w = z_1^\alpha$ this sector can be mapped onto a half plane.

\item[13.] 
If the circles are tangent to each other at the point $a$, the transformation $z_1 = 1/(z - a)$ will map the region between them onto a parallel strip, and a suitable exponential transformation maps the strip onto a half plane. 

\end{enumerate}

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\begin{enumerate}


\item[14.] 
A little more generally, the same method applies to a {\color{blue}circular triangle with two right angles}. 

\item[15.] 
In fact, if the third angle has the vertex $a$, and if the sides from $a$ meet again at $b$, the linear transformation $z_1 = (z - a)/(z - b)$ maps the triangle onto a {\color{blue}circular sector}. 

\item[16.] 
By means of a power this sector can be transformed into a {\color{blue}half circle}; the half circle is a wedge-shaped region which in turn can be mapped onto a {\color{blue}half plane}. 

\end{enumerate}

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\begin{enumerate}

\item[17.] 
In this connection we shall treat explicitly a special case which occurs frequently.

\item[18.] 
{\color{red}Let it be required to map the complement of a line segment onto the inside or outside of a circle. }

\item[19.] 
The region is a wedge with the angle $2\pi$; without loss of generality we may assume that the end points of the segment are $\pm 1$.

\item[20.] 
The preliminary transformation 
$$
z_1 = \frac{z+1}{z-1}
$$
maps the wedge on the full angle obtained by exclusion of the negative real axis. 

\end{enumerate}

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\begin{enumerate}

\item[21.] 
Next we define
$$
z_2 = \sqrt{z_1}
$$
as the square root whose real part is positive and obtain a map onto the right half plane. 

\item[22.] 
The final transformation
$$
w = \frac{z_2-1}{z_2+1}
$$
maps the half plane onto $|w| < 1$.


\end{enumerate}

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\begin{enumerate}

\item[23.] 
Elimination of the intermediate variables leads to the correspondence 
\begin{equation}
\begin{aligned}
z &= \frac{1}{2} \left( w+ \frac{1}{w} \right) \\
w &= z- \sqrt{z^2-1}.
\end{aligned}
\label{eq-16}
\end{equation}

\item[24.] 
The sign of the square root is uniquely determined by the condition $|w| < 1$, for 
$$
(z-\sqrt{z^2-1})(z+\sqrt{z^2-1})=1. 
$$ 
If the sign is changed, we obtain a mapping onto $|w| > 1$.
 

\end{enumerate}

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\begin{enumerate}


\item[25.] 
For a more detailed study of the mapping (\ref{eq-16}) we set $w = \rho e^{i\theta}$ and obtain 
\begin{equation*}
\begin{aligned}
x &= \frac{1}{2} \left( \rho + \frac{1}{\rho} \right) \cos\theta \\
y &= \frac{1}{2} \left( \rho - \frac{1}{\rho} \right) \sin\theta. 
\end{aligned}
%\label{eq-}
\end{equation*}

\item[26.] 
Elimination of $\theta$ yields
\begin{equation}
\frac{x^2}{ \left[\frac{1}{2}(\rho + \rho^{-1})\right]^2} 
+ \frac{y^2}{ \left[\frac{1}{2}(\rho - \rho^{-1})\right]^2} = 1
\label{eq-17}
\end{equation}
and elimination of $\rho$
\begin{equation}
\frac{x^2}{ \cos^2\theta } 
+ \frac{y^2}{ \sin^2\theta} = 1. 
\label{eq-18}
\end{equation}

\end{enumerate}

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\begin{enumerate}

\item[27.] 
{\color{red}Hence the image of a circle $|w| = \rho < 1$ is an ellipse with the major axis $\rho + \rho^{-1}$ and the minor axis $\rho^{-1}- \rho$.
}

\item[28.] 
The image of a radius is half a branch of a hyperbola. 

\item[29.] 
The ellipses (\ref{eq-17}) and the hyperbolas (\ref{eq-18}) are confocal. 

\item[30.] 
The correspondence is illustrated in Fig. 3-9.

\item[31.] 
Clearly, the transformation (\ref{eq-16}) allows us to include in our list of elementary conformal mappings the mapping of the outside of an ellipse or the region between the branches of a hyperbola onto a circular region. 

\item[32.] 
{\color{blue}It does not, however, allow us to map the inside of an ellipse or the inside of a hyperbolic branch.}

\end{enumerate}

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\begin{figure}[ht!]
\centering
\includegraphics[height=0.6\textheight, width=0.75\textwidth]{figure-3-9.png}
%\caption{Fig. 3-9. Mapping by $z=\frac{1}{2}(w+w^{-1})$ }
\end{figure}

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\begin{enumerate}

\item[33.] 
{\color{red}As a final and less trivial example we shall study the mapping defined by a cubic polynomial $w = a_0z^3 + a_1z^2 + a_2z + a_3$. }

\item[34.] 
The familiar transformation $z = z_1 - a_1/3a_0$ allows us to get rid of the quadratic term, and by obvious normalizations we can reduce the polynomial to the form $w = z^3 - 3z$.

\item[35.] 
The coefficient for $z$ is chosen so as to make the derivative vanish for $z = \pm 1$.

\end{enumerate}

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\begin{enumerate}

\item[36.] 
Making use of the transformation (\ref{eq-16}) we introduce an auxiliary variable $\zeta$ defined by
\begin{equation*}
z = \zeta + \frac{1}{\zeta}.
\end{equation*}

\item[37.] 
Our cubic polynomial takes then the simple form
\begin{equation*}
w = \zeta^3 + \frac{1}{\zeta^3}.
\end{equation*}

\item[38.] 
We note that each $z$ determines two values $\zeta$, but they are reciprocal and yield the same value of $w$. 

\item[39.] 
In order to obtain a unique $\zeta$ we may impose the condition $|\zeta| < 1$, but then the segment $(-2,2)$ must be excluded from the $z$-plane.

\end{enumerate}

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\begin{enumerate}

\item[40.] 
It is now easy to visualize the correspondence between the $z$- and $w$-planes. 

\item[41.] 
To the circle $|\zeta| = \rho < 1$ corresponds an ellipse with the semiaxes $\rho^{-1}\pm \rho$ in the $z$-plane, and one with the semiaxes $\rho^{-3} \pm \rho^3$ in the w-plane.

\item[42.] 
Similarly, a radius $\mathrm{arg}\, \zeta = \theta$ corresponds to hyperbolic branches in the $z$- and $w$-planes; the one in the $z$-plane has an asymptote which makes the angle $-\theta$ with the positive real axis, and in the $w$-plane the corresponding angle is $-3\theta$. 

\end{enumerate}

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\begin{enumerate}

\item[43.] 
The whole pattern of {\color{blue}confocal ellipses and hyperbolas} remains invariant, but when $z$ describes an ellipse $w$ will trace the corresponding larger ellipse three times. 

\item[44.] 
The situation is thus very similar to the one in the case of the simpler mapping $w = z^3$. 

\item[45.] 
For orientation the reader may again lean on Fig. 3-9.


\end{enumerate}

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\begin{enumerate}


\item[46.] 
{\color{red}For the region between two hyperbolic branches whose asymptotes make an angle $\le 2\pi/3$ the mapping is one to one.
}

\item[47.] 
We note in particular that the six regions into which the hyperbola $3x^2 -y^2 = 3$  and the $x$-axis divide the $z$-plane are mapped onto half planes, three of them onto the upper half plane and three onto the lower. 

\item[48.] 
The inside of the right-hand branch of the hyperbola corresponds to the whole $w$-plane with an incision along the negative real axis up to the point $-2$.

\end{enumerate}

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\begin{itemize}
\item  {\color{red}Question. 
(All mappings are to be conformal.)

Map the common part of the disks $|z| < 1$ and $|z - 1| < 1$ on the inside of the unit circle. Choose the mapping so that the two symmetries are preserved.
}

\item  


\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Map the region between $|z| = 1$ and $|z - \frac{1}{2}| = \frac{1}{2}$ on a half plane.
}

\item  


\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Map the complement of the arc $|z| = 1, y \ge 0$ on the outside of the unit circle so that the points at $\infty$ correspond to each other.

}

\item  



\end{itemize}

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\begin{itemize}
\item  {\color{red}Question. 
Map the outside of the parabola $y^2 = 2px$ on the disk $|w| < 1$ so that $z = 0$ and $z = -p/2$ correspond to $w = 1$ and $w = 0$.
(Lindel\"of.)
}

\item  


\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Map the inside of the right-hand branch of the hyperbola $x^2-y^2 = a^2$ on the disk $|w| < 1$ so that the focus corresponds to $w = 0$ and the vertex to $w = -1$.
(Lindel\"of.)
}

\item  


\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Map the inside of the lemniscate $|z^2-a^2| = \rho^2 (\rho > a)$ on the disk $|w| < 1$ so that symmetries are preserved. (Lindel\"of.)
}

\item  


\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Map the outside of the ellipse $(x/a)^2+(y/b)^2=1$ onto $|w| < 1$ with preservation of symmetries.
}

\item  


\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Map the part of the $z$-plane to the left of the right-hand branch of the hyperbola $x^2-y^2 = 1$ on a half plane. (Lindel\"of.)

Hint: Consider on one side the mapping of the upper half of the region by $w = z^2$, on the other side the mapping of a quadrant by $w = z^3-3z$.
}

\item  


\end{itemize}

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\begin{enumerate}

\item[1.] 
The visualization of a function by means of the corresponding mapping is completely clear only when the mapping is one to one. 

\item[2.] 
If this is not the case, we can still give our imagination the necessary support by the introduction of generalized regions in which distinct points may have the same coordinates. 

\item[3.] 
In order to do this it is necessary to suppose that points which occupy the same place can be distinguished by other characteristics, for instance a tag or a color. 

\item[4.] 
Points with the same tag are considered to lie in the same sheet or layer.


\end{enumerate}

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\begin{enumerate}

\item[5.] 
This idea leads to the notion of a Riemann surface. 

\item[6.] 
It is not our intention to give, in this connection, a rigorous definition of this notion. 

\item[7.] 
For our purposes it is sufficient to introduce Riemann surfaces in a purely descriptive manner. 

\item[8.] 
We are free to do so as long as we use them merely for purposes of illustration, and never in logical proofs.

\end{enumerate}

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\begin{enumerate}

\item[9.] 
The simplest Riemann surface is connected with the mapping by a power $w = z^n$, where $n > 1$ is an integer.

\item[10.] 
We know that there is a one-to-one correspondence between each angle $(k-1)(2\pi/n) < \mathrm{arg}\, z < k(2\pi/n)$, $k = 1, \cdots, n$, and the whole $w$-plane except for the positive real axis.


\item[11.] 
The image of each angle is thus obtained by performing a ``cut'' along the positive axis; this cut has an upper and a lower ``edge''.

\end{enumerate}

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\begin{enumerate}


\item[12.] 
Corresponding to the $n$ angles in the $z$-plane we consider $n$ identical copies of the $w$-plane with the cut. 

\item[13.] 
They will be the ``sheets'' of the Riemann surface, and they are distinguished by a tag $k$ which serves to identify the corresponding angle.

\item[14.] 
When $z$ moves in its plane, the corresponding point $w$ should be free to move on the Riemann surface. 

\item[15.] 
For this reason we must attach the lower edge of the first sheet to the upper edge of the second sheet, the lower edge of the second sheet to the upper edge of the third, and so on. 

\end{enumerate}

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\begin{enumerate}


\item[16.] 
In the last step the lower edge of the $n$th sheet is attached to the upper edge of the first sheet, completing the cycle.

\item[17.] 
In a physical sense this is not possible without self-intersection, but the idealized model shall be free from this discrepancy. 

\item[18.] 
The result of the construction is a Riemann surface whose points are in one-to-one correspondence with the points of the $z$-plane.

\item[19.] 
What is more, this correspondence is continuous if continuity is defined in the sense suggested by the construction.


\end{enumerate}

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\begin{enumerate}



\item[20.] 
The cut along the positive axis could be replaced by a cut along any simple arc from $0$ to $\infty$; the Riemann surface obtained in this way should be considered as identical with the one originally constructed. 

\item[21.] 
In other words, the cuts are in no way distinguished lines on the surface, but the introduction of specific cuts is necessary for descriptive purposes.

\item[22.] 
The point $w = 0$ is in a special position. 

\item[23.] 
It connects all the sheets, and a curve must wind $n$ times around the origin before it closes. 

\end{enumerate}

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\begin{enumerate}


\item[24.] 
A point of this kind is called a branch point. 

\item[25.] 
If our Riemann surface is considered over the extended plane, the point at $\infty$ is also a branch point.

\item[26.] 
In more general cases a branch point need not connect all the sheets; if it connects $h$ sheets, it is said to be of order $h-1$.


\end{enumerate}

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\begin{enumerate}



\item[27.] 
The Riemann surface corresponding to $w = e^z$ is of similar nature.

\item[28.] 
In this case the function maps each parallel strip $(k-1)2\pi < y < k\cdot 2\pi$ onto a sheet with a cut along the positive axis. 

\item[29.] 
The sheets are attached to each other so that they form an endless screw.

\item[30.] 
The origin will not be a point of the Riemann surface, corresponding to the fact that $e^z$ is never zero.


\end{enumerate}

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\begin{enumerate}



\item[31.] 
The reader will find it easy to construct other Riemann surfaces.

\item[32.] 
We will illustrate the procedure by consideration of the Riemann surface defined by $w = \cos z$.

\item[33.] 
A region which is mapped in a one-to-one manner onto the whole plane, except for one or more cuts, is called a {\color{blue}fundamental region}.

\item[34.] 
For fundamental regions of $w = \cos z$ we may choose the strips $(k-1)\pi < x < k\pi$. 

\end{enumerate}

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\begin{enumerate}


\item[35.] 
Each strip is mapped onto the whole $w$-plane with cuts along the real axis from $-\infty$ to $-1$ and from $1$ to $\infty$.

\item[36.] 
The line $x = k\pi$ corresponds to both edges of the positive cut if $k$ is even, and to the edges of the negative cut if $k$ is odd. 

\item[37.] 
If we consider the two strips which are adjacent along the line $x = k\pi$, we find that the edges of the corresponding cuts must be joined crosswise so as to generate a simple branch point at $w = \pm 1$.

\item[38.] 
The resulting surface has infinitely many simple branch points over $w = 1$ and $w = -1$ which alternatingly connect the odd and even sheets.


\end{enumerate}

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\begin{enumerate}



\item[39.] 
An attempt to illustrate the connection between the sheets is made in Fig. 3-10. 


\item[40.] 
It represents a cross section of the surface in the case that the cuts are chosen parallel to each other. 

\item[41.] 
The reader should bear in mind that any two points on the same level can be joined by an arc which does not intersect any of the cuts. 


\end{enumerate}

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\begin{figure}[ht!]
\centering
\includegraphics[height=0.35\textheight, width=0.8\textwidth]{figure-3-10.png}
%\caption{Fig. 3-10.  }
\end{figure}

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\begin{enumerate}



\item[42.] 
Whatever the advantage of such representations may be, the clearest picture of the Riemann surface is obtained by direct consideration of the fundamental regions in the $z$-plane. 

\item[43.] 
The interpretation is even simpler if, as in Fig. 3-11, we introduce the subregions which correspond to the upper and lower half plane.

\item[44.] 
The shaded regions are those in which $\cos z$ has a positive imaginary part. 

\end{enumerate}

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\begin{figure}[ht!]
\centering
\includegraphics[height=0.7\textheight, width=0.95\textwidth]{figure-3-11.png}
%\caption{Fig. 3-11 }
\end{figure}

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\begin{enumerate}


\item[45.] 
Each region corresponds to a half plane on which we mark the boundary points $1$ and $-1$. 

\item[46.] 
For any two adjacent regions, one white and one shaded, the half planes must be joined across one of the three intervals $(-\infty,-1), (-1,1), (1,\infty)$.

\item[47.] 
The choice of the correct junction is automatic from a glance at the corresponding
situation in the $z$-plane.

\end{enumerate}

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\begin{itemize}
\item  {\color{red}Question. 
Describe the Riemann surface associated with the function 
$$
w=\frac{1}{2}\left(z+\frac{1}{z}\right). 
$$
}

\item  

\end{itemize}

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\begin{itemize}
\item  {\color{red}Question. 
Same problem for $w = (z^2-1)^2$. 
}

\item  

\end{itemize}

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\begin{frame}{4.3. Elementary Riemann Surfaces. Exercise 3. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Same problem for $w = z^3 -3z$.
}

\item  

\end{itemize}

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